Random Sums via Conditional Expectation and Variance

Why Random Sums Need Conditioning

In many probability models, the total amount depends both on how many events occur and on how large each event is. For example, a hypothetical insurer’s aggregate claim amount depends on the random number of claims and the random size of each claim. A sum with a random number of terms cannot be handled by treating the count as though it were fixed.

Let $N$ be a nonnegative integer-valued random variable, and let $X_1,X_2,\ldots$ be identically distributed severities. Define the random sum

S=\sum_{i=1}^{N}X_i,

with $S=0$ when $N=0$. The key method is to condition on $N$: once the count is known to be $n$, the sum has exactly $n$ terms.

Definition: For the random-sum formulas in this tutorial, assume $N$ is independent of the entire sequence $X_1,X_2,\ldots$, and the $X_i$ are independent and identically distributed with finite mean $\mu_X=E[X_1]$ and finite variance $\sigma_X^2=\operatorname{Var}(X_1)$. Also assume $E[N]$ and $\operatorname{Var}(N)$ are finite. The severities need not be positive for the formulas to hold, although insurance amounts are ordinarily nonnegative.

The Mean and Variance Formulas

Given $N=n$, the random sum is $X_1+\cdots+X_n$. Independence of the severities gives conditional mean $n\mu_X$ and conditional variance $n\sigma_X^2$. Independence between $N$ and the severity sequence ensures that conditioning on $N=n$ does not change the severity distribution.

Theorem: Under the assumptions above,

E[S]=E[N]\,E[X_1]

\operatorname{Var}(S) =E[N]\operatorname{Var}(X_1) +\operatorname{Var}(N)\bigl(E[X_1]\bigr)^2.

The first variance term reflects variation in individual severities across the expected number of terms. The second reflects variation in the count: when the average severity is nonzero, uncertainty about how many terms are included adds uncertainty to the total.

Why the Formulas Hold

The mean follows by averaging the conditional mean over all possible counts. For the variance, use the law of total variance, which separates average variation within each value of $N$ from variation among the conditional means.

Condition on the count.
For $N=n$, independence and identical distribution give $E[S\mid N=n]=n\mu_X$ and $\operatorname{Var}(S\mid N=n)=n\sigma_X^2$. Thus $E[S\mid N]=N\mu_X$ and $\operatorname{Var}(S\mid N)=N\sigma_X^2$.

Average the conditional mean.
The law of total expectation gives $E[S]=E[E[S\mid N]]=E[N\mu_X]=E[N]\mu_X$.

Separate the two sources of variance.
The law of total variance gives $\operatorname{Var}(S)=E[\operatorname{Var}(S\mid N)]+\operatorname{Var}(E[S\mid N])$. Substituting the conditional quantities yields $E[N\sigma_X^2]+\operatorname{Var}(N\mu_X)=E[N]\sigma_X^2+\operatorname{Var}(N)\mu_X^2$.

This derivation shows why the count variance multiplies the square of the severity mean. It is the conditional mean $N\mu_X$ that changes when the count changes; the variance of that conditional mean is $\mu_X^2\operatorname{Var}(N)$.

Worked Example: A Count With Three Possible Values

A hypothetical repair plan produces $N$ service payments, where $P(N=0)=0.20$, $P(N=1)=0.50$, and $P(N=3)=0.30$. Each payment $X_i$, in dollars, is independently $1 or $7 with probability $0.50$ each, and the payment sequence is independent of $N$. Find the mean and variance of the total $S$.

Method. Calculate the count and severity moments, then substitute into the random-sum formulas. The common mistake avoided here is using only the average severity variance and ignoring variation in the count.

First calculate the count moments:

$$ E[N]=0(0.20)+1(0.50)+3(0.30)=1.40. $$

$$ E[N^2]=0^2(0.20)+1^2(0.50)+3^2(0.30) =0+0.50+2.70=3.20, $$

\operatorname{Var}(N)=E[N^2]-(E[N])^2 =3.20-(1.40)^2=3.20-1.96=1.24.

For one payment,

E[X_i]=1(0.50)+7(0.50)=4, \qquad E[X_i^2]=1^2(0.50)+7^2(0.50)=25.

\operatorname{Var}(X_i)=25-4^2=9.

Therefore,

E[S]=1.40(4)=5.60\text{ dollars},

\operatorname{Var}(S)=1.40(9)+1.24(4^2) =12.60+19.84=32.44\text{ dollars}^2.

As a check, the conditional mean is $4N$. Its expectation is $4(1.40)=5.60$, and its variance is $16(1.24)=19.84$. The average conditional variance is $9(1.40)=12.60$; adding the two variance components gives $32.44\text{ dollars}^2$, as above.

Worked Examples With Common Count Models

Worked Example: Poisson Count and Exponential Severity

In a hypothetical portfolio, the number of claims $N$ has a Poisson distribution with parameter $\lambda=2$. Claim amounts $X_i$, measured in thousands of dollars, are iid exponential with mean $5$ thousand dollars and independent of $N$. Find the mean and variance of the aggregate amount $S$.

Method. Use $E[N]=\operatorname{Var}(N)=\lambda$, and use the exponential severity’s mean and variance. This avoids the mistake of assuming that an exponential claim amount has variance equal to the square of its mean.

For an exponential distribution with mean $5$, the variance is $5^2=25$ in squared thousands of dollars. The count moments are

E[N]=2,\qquad \operatorname{Var}(N)=2.

Thus,

E[S]=2(5)=10\text{ thousand dollars},

\operatorname{Var}(S)=2(25)+2(5^2) =50+50=100\text{ (thousand dollars)}^2.

A check uses the conditional moments: given $N=n$, the aggregate mean is $5n$ and its variance is $25n$. Averaging the conditional variance gives $25(2)=50$; the variance of the conditional mean is $25(2)=50$. Their sum is $100\text{ (thousand dollars)}^2$.

Worked Example: Binomial Count and Two-Point Severity

A hypothetical equipment program has five independent units. Each unit independently generates a payment with probability $0.40$, so $N\sim\operatorname{Binomial}(5,0.40)$. Conditional on a payment occurring, its amount $X_i$, in thousands of dollars, is $2$ with probability $0.40$ and $5$ with probability $0.60$. Assume the payment amounts are iid and independent of the count. Find the aggregate mean and variance.

Method. Compute the binomial count moments and the two-point severity moments separately before applying the formula. This avoids confusing the probability of generating a payment with the distribution of its amount.

For the count,

E[N]=5(0.40)=2,\qquad \operatorname{Var}(N)=5(0.40)(0.60)=1.20.

For a payment amount,

E[X_i]=2(0.40)+5(0.60)=0.80+3.00=3.80 \text{ thousand dollars},

$$ E[X_i^2]=2^2(0.40)+5^2(0.60) =1.60+15.00=16.60, $$

\operatorname{Var}(X_i)=16.60-(3.80)^2 =16.60-14.44=2.16 \text{ (thousand dollars)}^2.

The aggregate moments are therefore

E[S]=2(3.80)=7.60\text{ thousand dollars},

\operatorname{Var}(S)=2(2.16)+1.20(3.80)^2 =4.32+17.328=21.648 \text{ (thousand dollars)}^2.

For a check, the average within-count variance is $E[N]\operatorname{Var}(X_i)=4.32$, while the variance of the conditional mean $3.80N$ is $3.80^2(1.20)=17.328$. Their sum is $21.648\text{ (thousand dollars)}^2$.

Common Mistake: Treating the Count as Fixed

A frequent error is to calculate $E[N]\operatorname{Var}(X_i)$ and report it as the whole variance. That quantity is only the average conditional variance. It omits the variation in the conditional mean caused by a random count.

Exam trap: In the first worked example, using only $E[N]\operatorname{Var}(X_i)$ gives $1.40(9)=12.60\text{ dollars}^2$. This is not the aggregate variance. The missing term is $\operatorname{Var}(N)(E[X_i])^2=1.24(4^2)=19.84\text{ dollars}^2$, so the correct variance is $12.60+19.84=32.44\text{ dollars}^2$.

The omitted term is zero only in special cases, such as a fixed count with $\operatorname{Var}(N)=0$, or a severity mean of zero. For nonnegative claim amounts with positive mean, a genuinely random count contributes additional variance. Also check the independence assumptions: if severities are dependent on the count or on one another, the conditional moments used in this derivation may not hold.

Key takeaway: For an independent count and iid severities, aggregate uncertainty has two components: average severity variance across the expected number of terms, plus count variance multiplied by squared mean severity. Condition on the count first, then combine the components.

Check Your Understanding

For each problem, assume the count is independent of the iid severity sequence and report monetary quantities in the stated units.

A hypothetical service plan has $N\sim\operatorname{Bernoulli}(0.30)$. Each payment has mean $8$ dollars and variance $16$ dollars squared. Find $E[S]$ and $\operatorname{Var}(S)$. (Answer: $E[S]=2.4$ dollars; $\operatorname{Var}(S)=18.24$ dollars squared.)
A hypothetical program has $P(N=0)=0.25$, $P(N=2)=0.50$, and $P(N=4)=0.25$. Each amount has mean $3$ dollars and variance $4$ dollars squared. Find the aggregate mean and variance. (Answer: $E[S]=6$ dollars; $\operatorname{Var}(S)=26$ dollars squared.)
A hypothetical claim count is Poisson with parameter $3$. Each claim amount has mean $2$ thousand dollars and variance $5$ squared thousand dollars. Find the aggregate mean and variance. (Answer: $E[S]=6$ thousand dollars; $\operatorname{Var}(S)=27$ squared thousand dollars.)
A hypothetical portfolio has $N\sim\operatorname{Binomial}(6,0.20)$. Each payment has mean $10$ dollars and variance $25$ dollars squared. Find the aggregate mean and variance. (Answer: $E[S]=12$ dollars; $\operatorname{Var}(S)=126$ dollars squared.)
A hypothetical count satisfies $P(N=1)=0.20$, $P(N=3)=0.50$, and $P(N=5)=0.30$. Each severity, in thousands of dollars, is $1$ with probability $0.50$ and $4$ with probability $0.50$. Find the aggregate mean and variance. (Answer: $E[S]=8$ thousand dollars; $\operatorname{Var}(S)=19.45$ squared thousand dollars.)

What You'll Learn